There is no such map. For convenience, I will use the upper half-plane $\mathbb{H}$ instead of the unit disk.
Suppose $f\colon \mathbb{H}\to\mathbb{C}$ is a proper holomorphic map.
Note first that, for each sequence $\{z_n\}$ in $\mathbb{H}$ that converges to a point $p$ on the real axis, the image sequence $\{f(z_n)\}$ must must converge to $\infty$ on the Riemann sphere. To prove this, observe that if $\overline{D}_R$ is the closed disk of radius $R$ in $\mathbb{C}$ centered at the origin, then $\overline{D}_R$ is compact, and hence $f^{-1}(\overline{D}_R)$ is a compact subset of $\mathbb{H}$. Then $\mathbb{C}-f^{-1}(\overline{D}_R)$ is an open neighborhood of $p$, so $z_n$ lies in the complement of $f^{-1}(\overline{D}_R)$ for all but finitely many $n$. This holds for all $R>0$, which proves the claim.
Now, since $f$ is a nonzero holomorphic function, the zeroes of $f$ form a discrete subset of $\mathbb{H}$. Since $f$ is proper, the set of zeroes must be compact, and therefore there are only finitely many zeroes. Let $D$ be any open disk centered on the real axis that does not contain any zeroes of $f$, and consider the holomorphic function $g(z) = 1/f(z)$ on $D\cap\mathbb{H}$.
Now, $g$ has the property that $g(z_n) \to 0$ for any sequence $z_n \in D\cap\mathbb{H}$ converging to a point on the real axis. Thus $g$ extends continuously to $D\cap\overline{\mathbb{H}}$ by setting $g(x) = 0$ for $x \in D\cap\mathbb{R}$. By the Schwarz reflection principle, $g$ now extends to a holomorphic function on all of $D$, which is nonsense since $g$ is zero on $D\cap\mathbb{R}$.
As the comments got too long, I will make them in an answer. Let $P$ a nonconstant polynomial in the plane. Then:
1: $P$ is proper as a map $P:\mathbb{C} \to \mathbb{C}$
2: If we restrict $P$ to a domain $D$ and $\Omega=P(D)$ is the image domain ($P$ is an open map being analytic nonconstant so $\Omega$ is open, while it is connected by continuity), then $P$ is proper as a map $P: D \to \Omega$ if and only if $P(\partial D) \cap \Omega =\emptyset$
Proof: $|P(z)| \to \infty$ as $|z| \to \infty$ hence the preimage of bounded sets by $P$ is bounded. But in the plane, $K$ compact iff $K$ is bounded and closed and obviously $P^{-1}(K)$ is then bounded by the above argument and closed by continuity, hence it is compact, hence $P$ is proper as a plane map.
For 2, we notice that if $K \subset \Omega$, $K$ compact, then $P^{-1}(K) \cap D=L$ is bounded since $P^{-1}(K)$ is so, but $L$ is only relatively closed in $D$. However if $P(\partial D) \cap \Omega =\emptyset$ and $x_n \in L, x_n \to x$, $P(x_n) \to P(x) \in K$ so $x \in \bar D, P(x) \in \Omega$. As $\bar D = D \cup \partial D$ and we cannot have $x \in \partial D$ by hypothesis, it follows $x \in D$, hence $x \in L$, hence $L$ closed, so compact.
Conversely, if $P$ restricted to $D$ is proper and $x \in \partial D, P(x)=y \in \Omega$, then $P(z)=(z-x)^nQ(z)+y, n \ge 1, Q(x) \ne 0$ hence the preimage of a small closed disc centered at $y$ and included in $\Omega$ has a component that is a a compact neighborhood of $x$ in the plane, hence its intersection with $D$ is not compact which contradicts the fact that $P$ is proper restricted to $D$.
A simple example to show that this happens is $P(z)=z(z-1)$ on the open unit disc. Clearly $0 \in P(D)$ but a small closed disc centered at zero has as preimage in the open disc, two components, one around zero compact, but one that is the intersection of a compact neighborhood of $1$ with the open unit disc, hence non-compact.
In the examples of the OP, $z \to z^n$ on the discs centered at the origin, obviously the boundary condition is satisfied, so the maps are proper when restricted too, but if we shift the domain to a disc containing $1$ but having another root of unit of order $n$ on the boundary, the restricted maps are not proper any more
Best Answer
Suppose $f$ is proper. So it must not be constant. For any $a\in\mathbb{C}$, $f^{-1}(a)$ is isolated by Identity Theorem. By properness $f^{-1}(a)$ is a finite set. The cardinality $n$, counted with multiplicity, of this set (I.e. The degree of $f$) is independent of $a$ by continuity. Actually $n$ is the number of zeros (counted with multiplicity) of $f(z)-a$. So $f$ is a polynomial of degree $n$.
Edit: 1. By Rouche's theorem, \begin{eqnarray}n=\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)-a}dz\end{eqnarray} for large enough contour $\gamma$ which encloses all the zeros. Wiggling $a$ does not change $n$.